24. Δ(델타)결선

24. Δ(델타)결선

$V_{L}=|\mathbf{V}_{ab}|=|\mathbf{V}_{bc}|=|\mathbf{V}_{ca}|$ (선전압)

$V_{p}=|\mathbf{V}_{an}|=|\mathbf{V}_{bn}|=|\mathbf{V}_{cn}|$ (상전압)

$V_{L}=\sqrt{3}V_{p}$, $\mathbf{V}_{ab}=\sqrt{3}V_{p}\angle30^{\circ}$

$\displaystyle\mathbf{I}_{AB}=\frac{\mathbf{V}_{ab}}{\mathbf{Z}_{p}}$, $\displaystyle\mathbf{I}_{BC}=\frac{\mathbf{V}_{bc}}{\mathbf{Z}_{p}}$, $\displaystyle\mathbf{I}_{CA}=\frac{\mathbf{V}}{\mathbf{Z}_{p}}$ ($\mathbf{Z}_{p}$는 유도성 임피던스)

3상 전류들의 크기: $I_{p}=|\mathbf{I}_{AB}|=|\mathbf{I}_{BC}|=|\mathbf{I}_{CA}|$

선전류의 크기: $I_{L}=|\mathbf{I}_{aA}|=|\mathbf{I}_{bB}|=|\mathbf{I}_{cC}|$

$I_{L}=\sqrt{3}I_{p}$

$\Delta$결선	상전압=선전압	선전류>상전류
Y결선	선전압>상전압	선전류=상전류

Y결선에서 역률각이 $\theta$이면 한 상이 흡수하는 전력은 $\displaystyle P_{p}=V_{p}I_{p}\cos\theta=V_{p}I_{L}\cos\theta=\frac{V_{L}}{\sqrt{3}}I_{L}\cos\theta$이고 전체 전력은 $P=3P_{p}=\sqrt{3}V_{p}I_{p}\cos\theta$이다.

$\Delta$결선에서 역률각이 $\theta$이면 한 상이 흡수하는 전력은 $P_{p}=V_{p}I_{p}\cos\theta=V_{L}I_{p}\cos\theta=V_{L}\frac{I_{L}}{\sqrt{3}}\cos\theta$이고 전체전력은 $P=3P_{p}=\sqrt{3}V_{p}I_{p}\cos\theta$이다.

부하	상전압	선전압	상전류	선전류
Y	$\mathbf{V}_{AN}=V_{p}\angle0^{\circ}$ $\mathbf{V}_{BN}=V_{p}\angle-120^{\circ}$ $\mathbf{V}_{CN}=V_{p}\angle-240^{\circ}$	$\begin{align*}\mathbf{V}_{AB}&=\mathbf{V}_{ab}=(\sqrt{3}\angle30^{\circ})\mathbf{V}_{AN}\\&=\sqrt{3}V_{p}\angle30^{\circ}\end{align*}$ $\begin{align*}\mathbf{V}_{BC}&=\mathbf{V}_{bc}=(\sqrt{3}\angle30^{\circ})\mathbf{V}_{BN}\\&=\sqrt{3}V_{p}\angle-90^{\circ}\end{align*}$ $\begin{align*}\mathbf{V}_{CA}&=\mathbf{V}_{ca}=(\sqrt{3}\angle30^{\circ})\mathbf{V}_{CN}\\&=\sqrt{3}V_{p}\angle-210^{\circ}\end{align*}$	$\displaystyle\mathbf{I}_{aA}=\mathbf{I}_{AN}=\frac{\mathbf{V}_{AN}}{\mathbf{Z}_{p}}$ $\mathbf{I}_{bB}=\mathbf{I}_{BN}=\frac{\mathbf{V}_{BN}}{\mathbf{Z}_{p}}$ $\displaystyle\mathbf{I}_{cC}=\mathbf{I}_{CN}=\frac{\mathbf{V}_{CN}}{\mathbf{Z}_{p}}$	$\displaystyle\mathbf{I}_{aA}=\mathbf{I}_{AN}=\frac{\mathbf{V}_{AN}}{\mathbf{Z}_{p}}$ $\displaystyle\mathbf{I}_{bB}=\mathbf{I}_{BN}=\frac{\mathbf{V}_{BN}}{\mathbf{Z}_{p}}$ $\displaystyle\mathbf{I}_{cC}=\mathbf{I}_{CN}=\frac{\mathbf{V}_{CN}}{\mathbf{Z}_{p}}$
$\Delta$	$\mathbf{V}_{AB}=\mathbf{V}_{ab}=\sqrt{3}V_{p}\angle30^{\circ}$ $\mathbf{V}_{BC}=\mathbf{V}_{bc}=\sqrt{3}V_{p}\angle-90^{\circ}$ $\mathbf{V}_{CA}=\mathbf{V}_{ca}=\sqrt{3}V_{p}\angle-210^{\circ}$	$\mathbf{V}_{AB}=\mathbf{V}_{ab}=\sqrt{3}V_{p}\angle30^{\circ}$ $\mathbf{V}_{BC}=\mathbf{V}_{bc}=\sqrt{3}V_{p}\angle-90^{\circ}$ $\mathbf{V}_{CA}=\mathbf{V}_{ca}=\sqrt{3}V_{p}\angle-210^{\circ}$	$\displaystyle\mathbf{I}_{AB}=\frac{\mathbf{V}_{AB}}{\mathbf{Z}_{p}}$ $\displaystyle\mathbf{I}_{BC}=\frac{\mathbf{V}_{BC}}{\mathbf{Z}_{p}}$ $\displaystyle\mathbf{I}_{CA}=\frac{\mathbf{V}_{CA}}{\mathbf{Z}_{p}}$	$\displaystyle\mathbf{I}_{aA}=(\sqrt{3}\angle-30^{\circ})\frac{\mathbf{V}_{AB}}{\mathbf{Z}_{p}}$ $\displaystyle\mathbf{I}_{bB}=(\sqrt{3}\angle-30^{\circ})\frac{\mathbf{V}_{BN}}{\mathbf{Z}_{p}}$ $\displaystyle\mathbf{I}_{cC}=(\sqrt{3}\angle-30^{\circ})\frac{\mathbf{V}_{CN}}{\mathbf{Z}_{p}}$

상당전력: Y부하: $\sqrt{3}V_{L}I_{L}\cos\theta$, $\Delta$부하: $\sqrt{3}V_{L}I_{L}\cos\theta$ ($\cos\theta$는 부하의 역률)

Y결선과 $\Delta$결선 사이의 평형 3상 부하의 관계: $\displaystyle\mathbf{Z}_{Y}=\frac{\mathbf{Z}_{\Delta}}{3}$ 

참고자료:

Engineering Circuit Analysis 8th edition, Hayt, Kemmerly, Durbin, McGraw-Hill